Question: Simplify and expand the following expression: $ \dfrac{r}{4r + 9}-\dfrac{r - 5}{4r - 4} $
Answer: In order to subtract expressions, they must have a common denominator. Get both fractions over a common denominator of $(4r + 9)(4r - 4)$ Multiply the first term by $\dfrac{4r - 4}{4r - 4}$ $ \begin{align*} \dfrac{r}{4r + 9} \times \dfrac{4r - 4}{4r - 4} & = \dfrac{(r)(4r - 4)}{(4r + 9)(4r - 4)} \\ & = \dfrac{4r^2 - 4r}{(4r + 9)(4r - 4)}\end{align*} $ Multiply the second term by $\dfrac{4r + 9}{4r + 9}$ $ \begin{align*} \dfrac{r - 5}{4r - 4} \times \dfrac{4r + 9}{4r + 9} & = \dfrac{(r - 5)(4r + 9)}{(4r - 4)(4r + 9)} \\ & = \dfrac{4r^2 - 11r - 45}{(4r - 4)(4r + 9)}\end{align*} $ Now we have: $ = \dfrac{4r^2 - 4r}{(4r + 9)(4r - 4)} - \dfrac{4r^2 - 11r - 45}{(4r - 4)(4r + 9)} $ Now both terms have a common denominator we can subtract the numerators: $ = \dfrac{4r^2 - 4r - (4r^2 - 11r - 45)}{(4r + 9)(4r - 4)} $ $ = \dfrac{4r^2 - 4r - 4r^2 + 11r + 45}{(4r + 9)(4r - 4)} $ $ = \dfrac{7r + 45}{(4r + 9)(4r - 4)}$ Expand the denominator: $ = \dfrac{7r + 45}{16r^2 + 20r - 36}$